“Give me six hours to chop down a tree and I will spend the first four sharpening the axe.”
– US President Abraham Lincoln
From the earliest days of recorded history, humans have dreamed of soaring into the sky. Itis believed that the first kite was invented in China in circa 1000 bc. This was followed bythe invention of the first rockets in China in circa 200 bc. Thirteen centuries later and about8000 km away in Italy, Renaissance artist Leonardo da Vinci drew several designs forflying machines, inspired by his anatomical study of birds. In 1783, two French men namedJean-Francois de Rozier and Marquis D’Arlandes made the first free aerial flight in aMontgolfier hot-air balloon. However, it wasn’t until 1903 that the era of manned, poweredflight was begun by two American brothers from Dayton, Ohio. Their Wright Flyer was
Ridiculous! If we were meantto fly we wouldbe born with
wings.
I heard that thosebrothers from Ohio
made a flyingmachine!
Ha Ha
Ha
‘‘Bird’s Eye View’’ The Wright brothers proved that great things can be
accomplished if you have a willingness to work hard, search for the important
facts, and dare to have an optimistic hope about the future.
the starting point for the design of manned aircraft which today have drawn together thefar expanses of this planet. However, what is often overlooked is that they also developedthe first operational aircraft engine to power it.
The word propulsion means driving forward. Therefore a propulsion system is amachine that produces thrust to drive an object forward. On air and space vehicles, athrusting force is produced by applying Newton’s 3rd Law of Action and Reaction. Theaction is accomplished by accelerating a working gas, normally by adding heat due tochemical combustion. The reaction to this acceleration (V ) produces a net thrust force(FN ) that propels an air vehicle forward (Figure 1.1).∑
F =∑
mV (1.1)∑F = T hrust − Drag = mV (1.2)
V = FN − D
m(1.3)
Thrust (FN ) is the driving force that propels an aircraft, helicopter, missile, or rocketforward. An aerospace propulsion system (engine or motor) is simply a device that con-verts power into thrust to propel an aerospace vehicle. Table 1.1 shows the most prevalenttypes of aerospace propulsion systems used today. Each of these engine types is coveredin this book.
1.1 Fundamental Equations
Most aerospace propulsion systems operate in a cyclic manner to produce a net workoutput from a supply of heat. Engines convert heat energy from available sources (suchas combustion of chemical fuels) into mechanical work, according to the laws of fluidmechanics and thermodynamics. Therefore, the performance of propulsion systems isgoverned by the conservation of mass, momentum, and energy. This section presents areview of these fundamental equations and terms.
• Can operate efficientlyat high supersonic Machnumbers (2.5–5.0)
• Generally cannotoperate at subsonicspeeds, so requires abooster rocket
Scramjet engines • Experimentalhypersonicvehicles
• Many difficult technicalchallenges, nooperational models yet
• Can operate athypersonic Machnumbers (5.0–15.0)
• Cannot operate atsubsonic or lowsupersonic speeds, sorequires a booster rocket
(Images courtesy of NASA, USAF, and National Museum of the USAF.)
1.1.1 Review of Terms
1.1.1.1 Systems
A system is an identifiable collection of matter that is under investigation. The mass orexpanse outside of a system is called the surroundings. Systems can be moveable orfixed. Types of systems are listed below:
• Isolated System – A system that is completely uninfluenced by its surroundings. Ithas a fixed mass, and no heat or work can cross the boundary (or control volume) ofthe system
• Closed System – One in which a fixed mass (control mass) is contained within aboundary at all times. A piston cylinder with closed intake and exhaust manifolds is anexample of such a system (Figure 1.2). The piston moves and compresses the air, butthe mass of air in the cylinder remains fixed. Energy (e.g., heat and work) can crossthe boundary, but with the manifolds closed, no mass can enter or leave the system.
Fundamentals 5
Piston
Boundary
Figure 1.2 Closed system
spool
Intake
mi
Compression Combustion Expansion Exhaust
•me
•
• •If mi = me, then it is a steady flow system
Figure 1.3 Open system
• Open Systems (also called flow systems) – A system where matter (mass) and energyare transferred across the boundary of a control volume (Figure 1.3). A steady flowopen system requires that the mass flows at entry, exit, and any intermediate pointwithin the boundary are the same. A gas turbine engine operating at steady speed infixed ambient conditions is an example of such a system.
1.1.1.2 Working Fluid
Most aerospace propulsion systems operate in a thermodynamic cycle that involves trans-ferring heat to and from a working fluid. Atmospheric air (or air mixed with combustiongases) is the predominant working fluid used in air-breathing propulsion systems. Air-breathing engines (such as gas turbine, ramjet, and scramjet engines) are open systemsthat draw in ambient air and use it as an oxidizer to burn fuel. However, the properties ofatmospheric air change with altitude. Because of this, these engines can operate only overa certain range of altitudes and velocities (Mach numbers) which correspond to differingatmospheric pressures, temperatures, and densities. This range is known as the engine’sflight envelope. Figure 1.4 shows approximate flight envelopes of aircraft with varioustypes of propulsion systems.
6 Aerospace Propulsion Systems
Altitude(kft (km)
15
10
20
30
0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0
Mach number
11.0
32
50
65
80
100SR-71
Blackbird
Commercialjets
Ramjets
Turbojet/turbofan limit
Turboprop limit
Piston engine limit
Helicopter limit
Scramjets
Hyper-X (X-43)
Militaryfighters
Figure 1.4 Flight envelopes of aircraft with different engine types
Rockets are non-air-breathing systems, so are not limited by altitude, so they can evenoperate in the vacuum of space. The working fluid of a chemical rocket is the combustiongases produced by combustion of its propellants.
The properties used to define the state of a working fluid are: pressure (P ), temperature(T ), volume (–V ), enthalpy (H ), density (ρ), internal energy (U ), and entropy (S). Sinceatmospheric air properties vary with altitude, weather, location, time, and other factors;the Standard Atmospheric Tables are typically used by engine designers as a standardreference [1]. Table A.1 in Appendix A gives the properties of the standard atmospherein SI units. Values such as temperature, pressure, density, and viscosity are given atdifferent altitudes. An altitude of 0 km represents the sea-level condition.
1.1.1.3 Work and Power
Work (W ) is generated when a force moves something in the direction it is being applied.It is a scalar quantity that is simply defined as the force applied to a body times the dis-placement of that body in that direction. If there is no displacement (or movement), thereis no work. It has the SI (Standard International) units newton-meter (N·m), which isalso called a joule (J). In Imperial units it is written in foot-pound (ft-lbf), Btu (Britishthermal units), or calories (cal). The generally accepted sign convention for work inter-action is that work done by a system (gain) is positive and work done on a system (loss)is negative.
Fundamentals 7
Sorry, I can’t help you with your schoolwork because I’m still working on thecomputer, as I have been for hours. But Dad! Since
you have notmoved from that
chair, myscience booksays that youhave not done
any work
‘‘Hard Work’’ The definition of mechanical work requires movement to
occur. The girl in this cartoon makes a great point, but she obviously did not
see her dad typing on the keyboard, moving his mouse or putting paper in
the printer.
Power (W ) is the rate of doing work. It is normally expressed in joules per second(J/s) or Watts (W ). In Imperial units it can be expressed as horsepower (hp), Btu per hour(Btu/hr) or in calories per second (cal/s).
1.1.1.4 Energy
Energy is a scalar physical quantity that describes the capacity of a system to producework. It is expressed in SI units as joules (J). It can exist in many different forms whichare often named after a related force. Some examples are kinetic, potential, thermal,mechanical, electrical, chemical, magnetic, and nuclear. The total energy (E) of a systemis the sum of all the forms of energy applicable to that system. The various forms of energycan be categorized into two groups: macroscopic and microscopic. The macroscopic formsof energy are those that a system has as a whole with respect to an outside frame ofreference. This would include the overall system’s kinetic and gravitational potentialenergies. The microscopic forms of energy are those that are related to the molecularstructure and activity of a system, independent of an outside frame of reference.
The sum of all microscopic energy is called the internal energy (U ). This includes thesum of the kinetic and potential energy of the molecules. Molecular kinetic energy consistsof atoms or electrons that may be vibrating, translating, rotating, or spinning (dependingon the phase of the matter). The portion of the internal energy related to the kineticenergy is called the sensible energy. The level of activity (velocity and momentum) ofthe molecules increases with temperature. As temperature increases, the sensible energy
8 Aerospace Propulsion Systems
increases causing the system to have a higher internal energy. Internal energy is also asso-ciated with the binding forces between molecules. A sufficient amount of energy can breakmolecular bonds causing a phase change (such as a liquid into a gas). The internal energyrelated to this phase change is called the latent energy. (A phase change process can occurwithout changing the chemical composition of the system.) The combination of the sensi-ble and latent internal energy of a system is commonly referred to as its thermal energy.
The energy associated with the atomic bonds in a system is called the chemical energy.During a combustion process some chemical bonds are formed while others are destroyed,changing the internal energy of the matter. Nuclear energy involves changes to the strongbonds within the nucleus of an atom. (An atom remains the same during a chemicalreaction, but changes in a nuclear reaction.) Mechanical energy is a form of energythat is converted into work by a mechanical device, such as a turbine. In a gas turbinepropulsion system, a propeller or compressor transfers mechanical work into a workingfluid by raising its pressure.
1.1.1.5 Heat
Energy can be transferred from one system to another as either work or heat. Heat (Q)is defined as the energy transferred between molecules of one system to those of anotherdue to a temperature difference. Like work, heat is expressed in SI units as joules (J).By convention, a positive value of heat indicates a heat gain to a system and a negativevalue indicates a heat loss.
Heat is transferred as it crosses the boundaries of one system into another. Once theheat transfers into a system it becomes part of the internal energy of that system. Heattransfer (Q) occurs by three mechanisms: conduction, convection, and radiation. Conduc-tive heat transfer occurs when an energetic substance (high thermal energy) is in contactwith a less energetic substance. Conduction can take place in solids, liquids, or gases. Insolids it occurs due to the combinations of vibrations of the molecules in their lattice andthe energy transport of free electrons. In liquids and gases, conduction takes place due tocollisions and diffusion of the molecules randomly in motion. Convective heat transferoccurs due to the combined effects of molecular motion (conduction) and the bulk motionof fluids (liquids or gases). An understanding of heat convected between a bounding sur-face and a moving fluid, when both are at different temperatures, is particularly importantin the analysis of propulsion systems (e.g., cooling). Radiation heat transfer is the energyemitted by matter in the form of electromagnetic waves (or photons). Unlike conductionor convection, radiation does not require the presence of an intervening medium, so itcan occur across the vacuum of space.
The thermal efficiency (ηth) of a heat engine is the ratio of its work output to the totalheat added into the system. If the total work is equal to the change of heat in the system,then this is expressed as Equation 1.4.
Thermal efficiency:
ηth = W
Qin= Qin − Qout
Qin= 1 − Qout
Qin(1.4)
Fundamentals 9
1.1.1.6 Cycles
A system undergoes a process when its state changes from one equilibrium condition toanother. If a system undergoes a number of processes so that it’s final state is the sameas its initial state (in all respects), then the system has undergone a cycle.
Figure 1.5 shows a pressure–volume plot (P − –V diagram) of an expansion process,involving non-repetitive translational motion of a piston cylinder. This is an example of anon-cyclic process from a high pressure/low volume state to a low pressure/high volume state.
Figure 1.6 shows an ideal gas turbine cycle. The numbered labels refer to different pro-cesses in the cycle. The last process (4 → 1) is not actually physically possible inside a gasturbine engine. The exhaust gases diffuse outside in the ambient air surroundings. However,since this is an open system, the engine operates in a cycle because ambient air continuouslyflows into the engine through its intake (which effectively restarts the cycle at point 1).
1.1.1.7 Isentropic Processes
Ideal processes are reversible. This means that the original state of the system can berestored, leaving no residual change in either the system or its surroundings. Reversible
(Not a cycle.) Work done in a process.
P
P A F1
2
Piston
V
Figure 1.5 Non-cyclic process
Heat addition
Expand through turbines
(Not possible)
Compression
2P
3
1 4
V
Figure 1.6 Cyclic process
10 Aerospace Propulsion Systems
Restraint
Pin
Pin
Gas
GasGas Heat
1
Work
2
3 4
Gas
Figure 1.7 Irreversible process
processes do not actually occur in nature. Consider the example illustrated in Figure 1.7.(1) A gas is held under pressure by a weightless piston that is pinned in place. (2) Whenthe pin is released, the pressurized gas pushes the piston upwards until a restraint stops it.Since the piston is weightless, there is no work done by the gas on the piston to move it.(3) In order to reverse this system to its original state a force (or work) must be applied topush the piston down and compress the gas. (4) Heat must also be transferred to the gasin order to restore its original internal energy. The addition of heat and work to restorethe system will change the surroundings. Since this process is not reversible, it is calledan irreversible process.
As the example shows, reversible processes do not occur in real systems because of thepresence of friction and finite temperature and pressure differences create irreversible con-ditions. Reversible systems are simply idealizations of an actual system. A work-producingdevice (e.g., turbine) will deliver the maximum amount of work if a reversible processis used. Similarly a work-consuming device (e.g., compressor) will consume a minimumamount of work when reversible processes are used. Therefore, reversible processes canbe considered as theoretical limits for corresponding real, irreversible processes. The morean actual process approximates a reversible process, the more efficient the design. There-fore reversible processes are often used to represent an ideal model of a system that canbe compared with the actual irreversible system.
A process in which no heat transfer occurs is called an adiabatic process. Normallyan adiabatic process can occur in only one of two ways: either the system is so wellinsulated that only a small (negligible) amount of heat can pass between the system andits surroundings, or both the system and its surroundings are at the same temperature. Anadiabatic, reversible process is called isentropic.
Fundamentals 11
The Second Law of Thermodynamics states that it is impossible to construct a systemthat will operate in a cycle, extract heat from a reservoir and do an equivalent amount ofwork on the surroundings. Thus the Second Law implies that if a system is to undergoa cycle and produce work, it must operate between two heat reservoirs of different tem-peratures. Therefore, an airplane driven by an engine extracting heat from the ambientair cannot exist, because it would have to reject heat to a lower temperature than itssurroundings in order to produce work. The fact that no system can truly be irreversible(and therefore isentropic) is shown by the Clausius inequality, which states:∮
δQ
T≤ 0 (1.5)
If no irreversibilities occur within the system then the cycle can be reversed. This is calledan internally reversible system, which is defined as:∮ (
δQ
T
)rev
= 0 (1.6)
In 1867, German physicist Rudolf Clausius defined the thermodynamic property, shownin Equation 1.5, as entropy (S) (Equation 1.6.) Entropy is expressed in SI units as J/K(or J/◦C). In Imperial units it can have units such as cal/◦R (or cal/◦F). Entropy per unitmass or specific entropy is represented by the symbol s.
dS =(
δQ
T
)rev
(1.7)
The Clausius inequality shows that entropy change (�S) of an isolated system during anirreversible process is always a positive quantity. In other words, some entropy is alwaysgenerated in an isolated, irreversible process. Entropy is constant in an isolated, reversibleprocess. Therefore, an isentropic process can be defined as an adiabatic process in whichthe entropy remains constant. This is known as the increase of entropy principle and issummarized as:
�S > 0 Irreversible processes (1.8)
�S = 0 Reversible (isentropic) processes (1.9)
�S < 0 Impossible (1.10)
Entropy is a useful property in analyzing propulsion systems, but unlike other proper-ties (such as temperature, pressure, or velocity) which have directly observable physicaleffects, entropy is a difficult concept for many people. It can be indirectly observed bymeasuring the losses incurred in other properties as a system undergoes a process. Entropycan be thought of as a measure of disorder, chaos, or randomness. Since no real, iso-lated system can undergo an isentropic process, the entropy or disorder of the universe isalways increasing.
Many systems used in engines such as compressors, turbines, nozzles and diffusers canbe ideally approximated by considering them to operate as an adiabatic process. Sincethe performance of these systems is optimized when irreversibilities (such as friction)are minimized, an isentropic (or ideal) model is often used to compare with the actual
12 Aerospace Propulsion Systems
irreversible process. This comparison is done by defining isentropic efficiencies thatcompare the actual performance to the idealized performance.
Clean up thismess! Your room
is total chaos!
But Momentropy alwaysincreases! So
this isirreversible.
‘‘Irreversible Mess’’ The Second Law of Thermodynamics says that pro-
cesses will naturally become more disordered. By intentionally adding heat
or work to a system we can bring more order to a particular attribute of
the system. However, even though one attribute of the system may become
more ordered, there will always be a net entropy gain (or increase of disorder)
due to other irreversibilities (such as friction or heat loss).
1.1.2 Equation of State for a Perfect Gas
The properties of gases (especially air) are very important in analyzing propulsion systems.An equation that relates the pressure, temperature, and density of a gas (or any substance)is called an equation of state. In 1662, Englishman Robert Boyle discovered that thepressure of a gas is inversely proportional to its volume, at constant temperature. Boyle’sLaw is shown in Equation 1.11.
P ∝ 1
–V(1.11)
In 1787, Frenchman Jacques Charles discovered that the volume of a gas is directlyproportional to its temperature, at constant pressure. In 1802, Joseph Louis Gay-Lussacpublished this finding (citing Charles’s unpublished work). This is now known as Charles’sLaw, shown in Equation 1.12.
T ∝ –V (1.12)
Combining the results of Boyle’s and Charles’s Laws, an equation for a perfect (or ideal)gas is defined in Equations 1.13, 1.14, and 1.15.
Fundamentals 13
Equation of state for a perfect gas:(Ideal Gas Law)
P –v = RT
P = ρRT
P –V = mRT
⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
P ≡ Pressure of gas (N/m2)
–v ≡ –V
m≡ Specific volume of gas (m3/kg)
T ≡ Temperature of gas (K)
ρ ≡ 1
–v≡ Density of gas (kg/m3)
–V ≡ Volume of gas (m3)
m ≡ Mass of gas (kg)
(1.13)
(1.14)
(1.15)
The constant of proportionality (R) is called the gas constant and is different for everygas. It can be determined from a universal gas constant (R◦) shown in Equation 1.16.The ideal gas equation of state is particularly useful for analysis of propulsion systems,because air is closely approximated to a perfect gas. For pure air, R is approximatelyequal to 287 J/(kg·K).
R = Ro
Mw
{Ro ≡ Universal gas constant [= 8.3145 kJ/(kmole·K)]
Mw ≡ Molecular weight of gas (kg/kmole)(1.16)
1.1.3 Law of the Conservation of Mass
The Law of the Conservation of Mass (also known as the continuity equation) appliedto a fluid crossing the boundary of a fixed control volume is shown schematically inFigure 1.8.
The conservation of mass integral equation is shown in Equation 1.17.
∂
∂t
∫∫∫controlvolume
ρ d–V +∫∫
controlsurface
ρ �V dA = 0 (1.17)
Total change of fluid massin the controlvolume per
unit time
Mass of flowentering the
control volumeper unit time
Mass of flowleaving the
control volumeper unit time
Figure 1.8 Law of the conservation of mass
14 Aerospace Propulsion Systems
If there is a steady flow through the control volume then the time rate of change of massin the control volume is zero or:
∂
∂t
∫∫∫controlvolume
ρ d–V = 0 (1.18)
So the steady flow continuity equation becomes:∫∫controlsurface
ρ �V dA = 0 (1.19)
If the velocity ( �V ) does not vary in either magnitude or direction across a cross-sectionalarea (A) that is normal to the flow direction, then:
m = ρVA = constant (1.20)
Example 1.1
Liquid oxygen (LOX) and liquid hydrogen (LH2) are steadily injected into a rocketthrust chamber at 8 kg/s and 1 kg/s respectively and ignited as shown in Figure 1.9.The combustion products are expelled from the rocket through a nozzle with a diam-eter of 25 cm.
Thrust chamber
Control volumeY
Nozzle
Exit Ae, VeX
O2
H2
Figure 1.9
If the density of the combustion gases is 0.175 kg/m3. Determine the exit velocity ofthe combustion gases.
Solution
The control volume of the rocket thrust chamber and nozzle is shown by the dottedline in Figure 1.9. Since the propellants are flowing at a steady rate, the conservationof mass equations are reduced to:
controlvolume
controlsurface
0, steady flow
Fundamentals 15
ρcombprod
AeVe︸ ︷︷ ︸mexit
= mO2 + mH2︸ ︷︷ ︸min
Ve = mO2 + mH2
Aeρcombprod
= mO2 + mH2
πd2
4
(ρcomb
prod
)
=8
kg
s+ 1
kg
sπ
4(0.25 m)2
(0.175
kg
m3
) = 1047.7m
s
1.1.4 Law of the Conservation of Linear Momentum
The conservation of linear momentum equation applied to a fixed control volume is shownschematically in Figure 1.10. The conservation of linear momentum integral equation isshown in Equation 1.21:
∑F = ∂
∂t
∫∫∫controlvolume
ρ �V d –V +∫∫
controlsurface
�V(ρ �V dA
)(1.21)
In this equation, �F is the summation of all the external forces acting on the controlvolume, which may include gravity, pressure forces, or viscous forces. If steady flow con-ditions exist, then the time rate of change of momentum in the control volume is zero or:
∂
∂t
∫∫∫controlvolume
ρ �V d–V = 0 (1.22)
So the steady flow momentum equation becomes:∑F =
∫∫controlsurface
�V(ρ �V dA
)(1.23)
Summation ofall external
forces actingon the
control volume
Time rate ofchange of linear
momentumin the
control volume
Net flow rateof linear
momentum out of the
control volume
Figure 1.10 Conservation of linear momentum
16 Aerospace Propulsion Systems
Example 1.2
A rocket motor burning on a test stand steadily exhausts 20 kg/s of combustion gases atan exit velocity of 750 m/s, as shown in Figure 1.11. The static pressure of the exhaustgases exiting the nozzle is 120 kPa. Assume an ambient air pressure of 101.3 kPa.Determine the force (or thrust) the rocket produces.
Control volume
Thrust chamber
Fn
Rx
Y
Nozzle
Ambient P0 = 101.3 kPa
Exhaust Ae = 0.02 m2
Ve = 750 m/s Pe = 120 kPaX
Figure 1.11
Solution
The external reaction force (Rx) which holds the rocket in place on the test stand isequal in magnitude but opposite in direction to the thrust force produced by the rocket.
FN = −Rx
The control volume encompassing the rocket and test stand is shown by the dottedline in Figure 1.11. Since the exhaust gas is flowing at a steady rate, the conservationof momentum equations reduce to:
controlvolume
controlsurface
0, steady flow
FN = mV + Ae (Pe − P0)
=
(20
kg
s
)(750
m
s
)1, 000
N
kN
+ (0.02 m2) (120 kPa − 101.3 kPa)
= 15.4 kNThis example shows how the thrust equation for a rocket is derived from the Law ofthe Conservation of Momentum.
Fundamentals 17
1.1.5 Law of the Conservation of Energy
The Law of the Conservation of Energy is also known as the First Law of Thermody-namics. For a system consisting of a fixed mass of particles, the Law of the Conservationof Energy of the system is shown schematically in Figure 1.12.
Total rate ofenergy transferinto a control
volume by heatand work transfer
Time rate ofchange of
energy in thecontrol volume
Total flow rateof energy outof the control
surface bymass flow
Figure 1.12 Conservation of energy
The total heat (Q) and work (W ) are defined in the energy equation as forms of energycrossing the control volume boundary of a system. Total work can include mechanicalwork, electrical work, or magnetic work. The total energy (E) includes all energy formsof the system at a given state. The forms of energy include internal energy (U ), whichis the random motion of molecules in the system; the potential energy (EPE); kineticenergy (EKE), due to the position and motion of the entire system; and storable forms ofenergy such as chemical energy or electrical (capacitance) energy. Each of these energyforms must be applied to a control volume encompassing the system to derive an energyintegral equation. The total energy per unit mass is symbolized by e and the total internalenergy per unit mass (or specific internal energy) is symbolized by u. If the systemhas only internal, potential, and kinetic energies, the integral equation is expressed asEquation 1.24.
d
dt(Q − W) = ∂
∂t
∫∫∫controlvolume
eρ d–V +∫∫
controlsurface
(u +
�V 2
2+ gz
)(ρ �V dA
)(1.24)
If the system consists of mass flows that cross the control volume boundaries, the totalwork (W ) can be written in two parts. The first part is the flow work. This is the workrequired to overcome the external pressure forces and drive the mass across the boundaries.The second part lumps together all other work crossing the boundaries such as mechanical(or shaft) work, viscous shear work, electrical work, and so on. The equation for the flowwork is shown in Equation 1.25.
Wflow = P�–V = P
ρ�m (1.25)
Therefore the flow work done per unit mass is:
wflow = P
ρ(1.26)
Under these conditions, the energy equation is normally written by combining the specificinternal energy (u) and the flow work per unit mass (wflow) into a single property, called
18 Aerospace Propulsion Systems
the specific enthalpy (h) defined in Equation 1.27.
h = u + P
ρ(1.27)
The energy equation in this form is shown in Equation 1.28.
d
dt
(Q − W ′) = ∂
∂t
∫∫∫controlvolume
eρd –V +∫∫
controlsurface
(h +
�V 2
2+ gz
)(ρ �V dA
)(1.28)
In this equation W ′ is the total work excluding the flow work (wflow).Another important parameter used in relation to the energy equation is the specific heat.
The specific heat of a solid or liquid is defined as the amount of heat required to raise a unitmass through a 1◦C temperature rise. When the volume is held constant, this known as theconstant volume specific heat (Cv). Since the specific internal energy (u) is a functionof only the temperature of a perfect gas, this can be expressed as shown in Equation 1.29.
Cv =[
∂u
∂T
]–V
(1.29)
When the pressure is held constant, this is known as the constant pressure specific heat(Cp). However, from the definition of enthalpy (h) in Equation 1.27, it can be seen thatfor a perfect gas (such as air):
dh = du + d
(P
ρ
)(1.30)
Since this is a perfect gas, Equation 1.14 can be substituted into Equation 1.30 to give:
dh = du + RdT (1.31)
Therefore the enthalpy of a perfect gas is also only a function of temperature, so theconstant pressure specific heat can be expressed as Equation 1.32.
Cp =[
∂h
∂T
]P
(1.32)
Rearranging Equation 1.31 gives:
R = dh
dT− du
dT= Cp − Cv (1.33)
Fundamentals 19
Also the ratio of specific heats (γ ) is defined as:
γ = Cp
Cv
(1.34)
Therefore:Cv
R= Cv
Cp − Cv
= 1Cp
Cv− 1
= 1
γ − 1(1.35)
and likewise:Cp
R= γ
γ − 1(1.36)
Table A.2 in Appendix A lists properties of gases at a reference temperature of25◦C (298 K).
A calorically perfect gas is a perfect gas with constant specific heats. If a calori-cally perfect gas is involved in a thermodynamic process between two equilibrium states,then the specific internal energy or enthalpy difference can be determined by integratingEquations 1.29 and 1.32.
�u = u2 − u1 =2∫
1
Cv dT = Cv (T2 − T1) (1.37)
�h = h2 − h1 =2∫
1
Cp dT = Cp (T2 − T1) (1.38)
Another useful thermodynamic equation for a pure substance is:
T ds = dh − dP
ρ(1.39)
Therefore the entropy change of a perfect gas is:
s2 − s1 =2∫
1
Cp
dT
T− R ln
(P2
P1
)(1.40)
For an isentropic process �s = 0, therefore:
2∫1
Cp
dT
T= R ln
(P2
P1
)(1.41)
Cp
Rln
(T2
T1
)= ln
(P2
P1
)(1.42)
20 Aerospace Propulsion Systems
Example 1.3
Air enters an adiabatic compressor of a turbojet engine at 70 kg/s and at a tempera-ture (T1) of 30◦C, as shown in Figure 1.13. It flows steadily through the compressorwith no change in velocity and exits at a temperature (T2) of 350◦C. Assume thatthe constant pressure specific heat (Cp) of air is 1.005 kJ/(kg·K). Determine the min-imum power that must be generated by a turbine in order to drive the compressor atthese conditions.
Compressor Turbine Y
X1 2
Figure 1.13
Solution
The control volume around the compressor section is shown by the dotted line inFigure 1.13. The energy equation for steady flow is:
controlvolume
controlsurface
0, adiabatic0, steady flow No height change
No velocity change
Therefore the power required by the compressor to pressurize the air (and therebyalso increase its temperature) is equal to the minimum power required by the turbineto drive it.
1.2.1 Isentropic Relationship between Temperature and Pressure
Isentropic equations are often used in the analysis of perfect gas turbine engines and rocketmotors. For a simple, stationary compressible system the state postulate specifies that thestate of a system is completely specified by two independent, intensive (independent ofmass) properties, such as pressure (P ) and temperature (T ). (For a complete descriptionof the state postulate, see [2].) This means that the properties of a unit mass of a gas(a single-phase system) can be determined from knowledge of just two independent,intensive properties. However, if the gas is in motion, three intensive properties arerequired, such as velocity (V ), pressure (P ), and temperature (T ). The tables in AppendixB list how several property ratios vary ideally (isentropically) with Mach number. Theseratios are determined from isentropic equations, derived by first applying the conservationof energy equation to a closed stationary system (fixed mass) containing a compressibleworking fluid (such as air). An internally reversible process involving this closed systemis expressed in Equation 1.43.
dQ − dW = dU (1.43)
Since:dQ = T dS (1.44)
dW = Pd –V (1.45)
dU = Cv dT (1.46)
Substituting Equations 1.44, 1.45, and 1.46 into Equation 1.43 and recognizing that foran isentropic process T ds = 0, gives the following expression:
T ds = Cv dT + Pd –v = 0 (1.47)
Therefore:
dT = −Pd –v
Cv
(1.48)
Rearranging Equation 1.13 (for a perfect gas):
T = P –v
R(1.49)
Differentiating Equation 1.49 with respect to temperature gives:
dT = 1
R(Pd –v + –v dP) (1.50)
Substituting Equation 1.50 into Equation 1.48 gives:
−Pd –v
Cv
= 1
R(Pd –v + –v dP) (1.51)
Cv
R(Pd –v + –v dP) + Pd –v = 0 (1.52)
22 Aerospace Propulsion Systems
Since Equation 1.35 states:
Cv
R= 1
γ − 1
Therefore:1
γ − 1(P d –v + –v dP) + Pd –v = 0 (1.53)
Pd –v + –v dP + (γ − 1)P d –v = 0 (1.54)
Pd –v + –v dP + γPd –v − Pd –v = 0 (1.55)
dP
P+ γ
d–v
–v= 0 (1.56)
ln(P ) + γ ln(–v) = 0 (1.57)
These equations can be simplified by using the following logarithmic identities:
log(xn) = n log(x) (1.58)
log(xy) = log(x) + log(y) (1.59)
By applying these identities, Equation 1.57 can be rewritten as:
ln(P ) + ln(–vγ) = ln
(P –vγ
) = 0 (1.60)
Therefore:P –vγ = constant = C (1.61)
Since by the Ideal Gas Law (Equation 1.13):
–v = RT
P(1.62)
P –vγ = P
(RT
P
)γ
= T γ Rγ
P (γ−1)= constant = C (1.63)
Tγ
1 Rγ
P(γ−1)
1
= Tγ
2 Rγ
P(γ−1)
2
(1.64)
T1
T2=[P1
P2
] (γ−1)
γ
(1.65)
The isentropic relationship between temperature and pressure in a gas turbine engineis commonly illustrated by a T–S diagram of the engine cycle (Figure 1.14). The T–Sdiagram is generally preferred to the P– –V diagram for illustrating propulsion analysis ofgas turbine, ramjet, or scramjet engines. The ideal (or isentropic) processes representingthe compressor and turbine are shown by the vertical arrows and labeled with a primesymbol (′). The actual (or non-isentropic) processes are represented by dashed arrows.
Fundamentals 23
Isen
trop
ic
Isen
trop
ic
Com
pres
sor
Combustor
Tur
bineA
ctual
Act
ual
1
4′
2′
2
T
4
3
Pt3
Pt4
Pt1
Pt2
Tt 3
Tt 2
T ′t 2
S
Figure 1.14 T –S Diagram for a gas turbine engine
1.2.2 Isentropic Relationships with Specific Volume
It is also useful to develop isentropic expressions that relate the specific volume (–v) topressure (P ) and temperature (T ). Starting with the definition of enthalpy (h), shown inEquation 1.30:
dh = du + d
(P
ρ
)= du + d(P –v) (1.66)
dh = du + P d–v + –v dP (1.67)
Rearranging Equation 1.48:
Cv dT = −P d –v (1.68)
Substituting Equation 1.67 into Equation 1.68 gives:
0 = (dh − P d–v − –v dP) + P d–v (1.69)
0 = dh − –v dP (1.70)
dh = –v dP (1.71)
Since also (Equation 1.32):
dh = Cp dT (1.72)
then:
–v dP = Cp dT (1.73)
24 Aerospace Propulsion Systems
Dividing Equation 1.68 by Equation 1.73 results in the following equation:
−Pd –v
–v dP= Cv
Cp
(1.74)
dP
P= −Cp
Cv
d–v
–v= −γ
d–v
–v(1.75)
Integrating this equation between points 1 and 2 gives:
P2∫P1
dP
P= −γ
–v2∫–v1
d–v
–v(1.76)
ln
(P2
P1
)= −γ ln
(–v2
–v1
)(1.77)
ln
(P2
P1
)= ln
(–v2
–v1
)−γ
(1.78)
P2
P1=(
–v2
–v1
)−γ
=(
–v1
–v2
)+γ
(1.79)
Recall from Equation 1.13 for a perfect gas:
P = RT
–v
This can be substituted in to give a relationship with temperature.
P2
P1=
RT2
–v2RT1
–v1
=(
–v1
–v2
)γ
(1.80)
T2
T1
(–v1
–v2
)=(
–v1
–v2
)γ
(1.81)
T2
T1=(
–v1
–v2
)−1 ( –v1
–v2
)γ
(1.82)
T2
T1=(
–v1
–v2
)γ−1
(1.83)
Fundamentals 25
P n = ∞
n = 0n = 1 (isothermal) T = constant
n = g (isentropic)
Figure 1.15 Polytropic expression definition
1.3 Polytropic Processes
In the previous section, relations were derived using isentropic processes. These idealrelationships are often used to compare actual processes by use of isentropic efficiencies(Figure 1.14). Isentropic efficiencies will be defined and described in greater detail insubsequent chapters of this book. These efficiencies will be used as a measure of theirreversibility of turbomachinery processes. However, there are other propulsion processesthat approach the isentropic law. Polytropic processes follow laws that form the relation:
P –vn = constant = C (1.84)
Where n lies between 0 and γ . Equation 1.84 is used to create P– –V diagram plots forthree different profiles shown in Figure 1.15.
The different curves shown in this plot (created by changing the value of n) illustratealternative methods of defining efficiencies, called polytropic efficiencies. Figure 1.15shows that the isothermal (or constant temperature) process can be considered as a specialcase of the polytropic process when n = 1. Isentropic processes can be considered aspecial polytropic case, when n = γ . For simplicity, this book deals only with isentropicefficiencies. These are defined as the ratio of the ideal (isentropic) work to the actual workfor given pressure ratios. Other references make more use of polytropic efficiencies [3].Polytropic efficiencies are defined as the ratio of the ideal work to the actual work for adifferential pressure change.
1.4 Total (or Stagnation) Properties
When a fluid in motion is isentropically brought to rest a temperature and pressure riseoccurs. The properties of a fluid at this stagnation point are called stagnation proper-ties (Figure 1.16). In the absence of hydrostatic pressures (i.e., the elevation effects offluid weight on pressure), stagnation properties are equivalent to total properties. Totalproperties (e.g., Pt , Tt , ht ) are useful as a reference state for compressible flow in propul-sion systems.
26 Aerospace Propulsion Systems
Stagnation point (flow velocity = 0)(Tt, Pt)
T, P
Figure 1.16 Definition of total (or stagnation) properties
A definition for the total temperature can be derived by applying the energy equationto a unit mass flow in a duct where:
• there is no change in potential energy• flow is adiabatic (Q = 0)• no work is done (W = 0).
Applying the Law of the Conservation of Energy gives:
u1 + P1 –v1 + V 21
2= u2 + P2 –v2 + V 2
2
2(1.85)
CvT1 + RT1 + V 21
2= CvT2 + RT2 + V 2
2
2(1.86)
Substituting Equation 1.33 into this equation gives:
(Cp − R)T1 + RT1 + V 21
2= (Cp − R)T2 + RT2 + V 2
2
2(1.87)
T1 + V 21
2Cp
= T2 + V 22
2Cp
(1.88)
The sum of the terms on the two sides of this equation are identical. Therefore thesesummations can be defined as a new term, as shown in Equations 1.89 and 1.90.
Tt1 = Tt2 (1.89)
The term Tt in Equation 1.89 is the total (or stagnation) temperature (Figure 1.16). There-fore the total temperature is defined as:
Total (or stagnation) temperature:
Total temperature = Static temperature + Dynamic temperature
Tt = T + V 2
2Cp
(1.90)
Fundamentals 27
In this equation, the dynamic temperature is the temperature equivalent of the kineticenergy (V 2/2Cp) of the flow. The total to static pressure ratio (Pt/P ) can be related tothe total to static temperature ratio (Tt/T ) by inspection of Equation 1.65:
Pt
P=(
Tt
T
) γ
γ−1(1.91)
The total enthalpy per unit mass (ht ) for a perfect gas with constant specific heats istherefore defined as:
ht − h = Cp (Tt − T ) (1.92)
Substituting Equation 1.90 into Equation 1.92:
ht = h + V 2
2(1.93)
It is often inconvenient to use the velocity of the gas in aerospace propulsion analysis.The Mach number (M) is a non-dimensional parameter that is often used in place of thevelocity to describe the state of a flowing gas. The Mach number is defined as the ratioof the velocity (V ) over the velocity of sound (a).
Mach number:
M = V
a(1.94)
The velocity of sound (a) for a perfect gas is derived by encompassing a sound expan-sion wave within a control volume and applying the Law of the Conservation of Massand the Law of the Conservation of Momentum to it. Combining these two equationsresults in an equation for the velocity of sound (for a weak compression wave), shownin Equation 1.95 (the derivation is given in [4]).
a2 =(
∂P
∂ρ
)∣∣∣∣s
= dP
dρ(1.95)
The velocity of sound for a perfect gas is simplified from this expression by applyingEquation 1.79, which states:
P2
P1=(
–v1
–v2
)γ
=(
ρ2
ρ1
)γ
(1.96)
or:
P1
ργ
1
= P2
ργ
2
= P
ργ= constant = C (1.97)
28 Aerospace Propulsion Systems
Solving this equation for P and substituting this back into Equation 1.95 yields:
a2 = dP
dρ= C
d(ργ )
dρ= γCργ
ρ(1.98)
Substituting the value of P (found in Equation 1.97) into Equation 1.98 results in:
a2 = γP
ρ(1.99)
Simplifying this equation and substituting for (P/ρ ) by using the equation of state for
a perfect gas (Equation 1.14), results in:
Velocity of sound (perfect gas):
a =√
γP
ρ=√
γRT (1.100)
The total to static ratios of temperature and pressure can be written in terms of Machnumber. Combining Equations 1.99 and 1.100 together gives:
V 2 = M2a2 = M2 (γRT ) (1.101)
Substituting this equation into Equation 1.91 yields:
Tt
T= 1 + M2γR
2Cp
(1.102)
Recall from Equations 1.33 and 1.34:
R = Cp − Cv and γ = Cp
Cv
Combining these equations together gives:
R
Cv
= Cp
Cv
− 1 = γ − 1 (1.103)
R = Cv (γ − 1) (1.104)
Substituting Equation 1.104 into Equation 1.102 gives:
Tt
T= 1 + M2γCv (γ − 1)
2Cvγ(1.105)
Canceling out terms in this equation gives an equation relating total to static temperatureratio and Mach number.
Tt
T= 1 + M2 (γ − 1)
2(1.106)
Fundamentals 29
Substituting Equation 1.91 into Equation 1.106 gives an equation that relates the total tostatic pressure ratio and the Mach number.
Pt
P=[
1 + M2 (γ − 1)
2
] γ
γ−1(1.107)
Like Mach number (M), there are several other non-dimensional parameters that are usefulin engine analysis. One commonly used parameter is called the X -function. It is derivedby applying the Law of the Conservation of Mass to describe a steadily, flowing gas ina parallel duct (Equation 1.108).
m = ρAV (1.108)
Rearranging the equation of state for a perfect gas (Equation 1.14):
ρ = m
–V= P
RT(1.109)
Rearranging Equation 1.101:
V = M√
γRT (1.110)
Substituting Equation 1.109 and 1.110 into Equation 1.108 gives:
m = PAM√
γ√RT
(1.111)
Rearranging Equation 1.106 gives:
T = Tt(1 + γ + 1
2M2) (1.112)
Rearranging Equation 1.107 gives:
P = Pt(1 + γ + 1
2M2) γ
γ−1
(1.113)
Substituting Equations 1.112 and 1.113 into Equation 1.111 gives:
m = PtAM√
γ√RTt
[1 + (γ − 1) M2
2
]−(γ+1)
2(γ−1)
(1.114)
A portion of Equation 1.114 is defined as the non-dimensional mass flow rate or simplyas the X -function (X) (shown in Equation 1.115). The X-function is a useful term to aid
30 Aerospace Propulsion Systems
analysis of fluid flows because it is a unique function of Mach number in a caloricallyperfect gas.
X-function (non-dimensional mass flow rate):
X = M√
γ
[1 + γ − 1
2M2]−(γ+1)
2(γ−1)
(1.115)
Substituting the definition of the X-function (Equation 1.115) into Equation 1.114 gives:
m = PtAX√RTt
(1.116)
Therefore the X-function can also be expressed by rearranging Equation 1.116,which gives:
X-function (non-dimensional mass flow rate):
X = m√
RTt
APt
(1.117)
The X-function is often used in the analysis of propulsion systems. Likewise a Y -functionand Z -function can also be used. These are defined as:
X = YZ (1.118)
The Y - and Z-functions are shown in Equations 1.119 and 1.120.
Y-function (non-dimensional specific internal thrust reciprocal):
Y =M
√γ
√1 +
(γ − 1
2
)M2
1 + γM2(1.119)
Z-function (non-dimensional internal thrust):
Z =(1 + γM2
)[
1 + γ − 1
2M2] γ
γ−1
(1.120)
Fundamentals 31
1.5 Isentropic Principles in Engine Components
1.5.1 Ducts
Steady, isentropic flow through a frictionless duct is one of the simplest fluid dynamicssystems to define and analyze. No work can be extracted from a duct. If the flow isisentropic (and therefore adiabatic) there is no heat transfer (in or out), and the energyequation becomes:
Q − W = 0 (1.121)
Since there is no heat transfer, the total temperature is constant (�Tt = 0). For steadyflow (�m = 0) in a frictionless duct the total pressure is also constant (�Pt = 0). If africtional duct is considered (non-isentropic), there is a loss in total pressure because of thechange in entropy (�s) caused by frictional losses. This can be seen in Equation 1.122.
�s = �S
m= −R ln
(Pt2
Pt1
)(1.122)
Since the change in entropy (�s) cannot fall in value (Equation 1.8), then there mustbe a loss in total pressure (Pt2 < Pt1) in a frictional duct. But in this simple case, flowthrough the duct is non-frictional and isentropic (�s = 0). Therefore:
AX = m√
RTt
Pt
= constant = C (1.123)
The only forces acting on the gas in this duct are pressure forces (the duct exerts no axialforce). So applying Newton’s Second Law gives:
A (P1 − P2) = m (V2 − V1) (1.124)
ormV1 + AP1 = mV2 + AP2 (1.125)
Therefore at any plane in the duct:
mV + AP = constant (1.126)
This constant, known as the internal stream thrust (Fint ), is expressed as:
Fint = mV + AP (1.127)
Rearranging the equation for mass flow derived from the Law of the Conservation ofMass (Equation 1.20) gives:
m = PAV
RT(1.128)
Substituting Equation 1.128 into Equation 1.127 results in:
Fint = PAV 2
RT+ AP = PA
(V 2
RT+ 1
)(1.129)
32 Aerospace Propulsion Systems
Substituting the Mach number definition (Equation 1.94) for velocity into Equ-ation 1.128 gives:
Fint = PA(γM2 + 1
)(1.130)
Finally substituting Equation 1.107 into this equation gives:
Fint = PtA(γM2 + 1
)(
1 + M2(γ − 1)
2
) γ
γ−1
(1.131)
Substituting the Z−function into this equation gives:
Steady frictionless flow in a parallel duct:
Fint = PtAZ (1.132)
This equation reveals the reason that the Z−function is known as the non-dimensionalinternal thrust, since:
Z = Fint
PtA(1.133)
The Law of the Conservation of Mass or continuity equation (Equation 1.19) for one-dimensional steady flow through a varying area duct can be written as:
(ρ + dρ) (A + dA) (V + dV ) − ρAV = 0 (1.134)
dρ
ρ+ dA
A+ dV
V= 0 (1.135)
Applying the steady flow, linear momentum equation (Equation 1.23) for this same controlvolume gives the equation:
pA +(
P + dP
2
)dA − (P + dP ) (A + dA) = (ρAV ) dV (1.136)
dP + ρV dV = 0 (1.137)
Multiplying Equation 1.135 by (ρV 2) and rearranging gives:
ρV dV = −dρV 2 − ρV 2 dA
A(1.138)
Substituting Equation 1.138 into Equation 1.137 results in:
dP + ρV 2[−dρ
ρ− dA
A
]= 0 (1.139)
Fundamentals 33
This equation is simplified by solving the general definition of the velocity of sound(Equation 1.95) for dρ and then substituted into Equation 1.137, resulting in:
dP + ρV 2(
− dP
ρa2− dA
A
)= 0 (1.140)
Further simplification of Equation 1.140 gives a relationship for isentropic flow in avarying duct or channel.
dP(1 − M2) = ρV 2 dA
A(1.141)
Equation 1.141 is significant because it shows the effect that Mach number has on flowinside a varying area duct or channel (Figure 1.17). For subsonic flow (M < 1), it can be
seen from this equation that a decrease in area (dA) (called a converging duct), results ina decrease in pressure (dP ) and an increase in velocity (dV ). Conversely an increase inarea (diverging duct) increases the pressure and decreases the velocity.
For supersonic flow (M > 1), the opposite trends occur. These results show that sub-sonic flow cannot be accelerated to supersonic flow in a converging duct or nozzle. A con-vergent–divergent (condi) duct or nozzle must be used to achieve this (see Section 1.5.4).
1.5.2 Turbomachinery
Turbomachinery (such as compressors or turbines) are designed to transfer work (notheat). Compressors are used to increase the pressure of a flow while turbines are used toextract work (or energy) from the flow. In a gas turbine engine, energy extracted fromturbines is used to power the compressors. The power required to drive the compressoris determined by the energy equation (derived in Example 1.3):
Wc = mCp (Tt2 − Tt1) (1.142)
Ideally, the temperature ratio in compressors and turbines is the minimum associated withthe pressure changes in the devices. So ideal turbomachinery can be considered isentropicand Equation 1.65 can be applied (but for total pressure and temperature ratios), as shownin Equation 1.143.
Tt1
Tt2=[Pt2
Pt1
] γ−1γ
(1.143)
Actual turbomachinery has irreversibilities due to friction on all of the wetted surfaces.However, it is difficult to account for these irreversibilities. Instead ideal (isentropic) prop-erties can be used as a basis for the engine’s performance analysis. Isentropic efficiencies(η) are then used to determine actual properties from the ideal. (This will be discussed inmuch greater detail in subsequent chapters.)
1.5.3 Combustion Chambers (Combustors)
Combustion chambers (or thrust chambers in rockets) generate heat by greatly increasingthe temperature of the flow. For gas turbine engines and rockets, this is generally done byburning fuel or propellant. Combustion chambers are designed to be steady flow devices,so they are essentially ducts with the capacity for heat addition. No work can be extracted.The heat produced by the combustor is therefore:
Q = mCp (Tt2 − Tt1) (1.144)
1.5.4 Nozzles
Two types of nozzles are primarily used in aerospace propulsion systems (shown inFigure 1.18). A convergent nozzle is shaped to have a continuously decreasing cross-sectional area in the flow direction. A convergent–divergent (or condi) nozzle is aconvergent nozzle followed by a divergent nozzle (continuous increase in cross-sectional
Fundamentals 35
m•
m•
Convergent nozzle Condi nozzle
i e ei
Figure 1.18 Types of nozzles
area in the flow direction). The condi nozzle was originally invented by a Swedish engineernamed Karl Gustaf de Laval for use in steam turbines and is therefore also known as thede Laval nozzle.
(The symbol ‘∗’ is generally used as a label to indicate the location of the throat,which is the point between the convergent and divergent portions of a condi nozzle. Acondi nozzle is optimally designed so that M∗ = 1.0 at the throat.)
1.5.4.1 Convergent Nozzles
Isentropic flow through a convergent nozzle can be best understood by examining a nozzlewith a constant chamber pressure (Pc) and applying decreasing back pressures (pointsA → D) on it (Figure 1.19). If the nozzle is exhausting gases into the atmosphere, thisback pressure is equal to the ambient pressure (P0). Therefore a continuous decrease inthe ambient back pressure is equivalent to climbing in altitude (Table A.1, Appendix A).
Point A illustrates a limiting case where the back pressure equals the chamber pressure(P0 = Pc), so there is no mass flow through the nozzle. As the ambient back pressure islowered to point B and beyond (P0 < Pc), the static pressure through the nozzle decreasesand the mass flow through the nozzle increases. The Mach number at the nozzle exit planealso increases. Under these conditions, the static pressure of the flow exiting the nozzleis equal to the ambient back pressure (Pe = P0). This is called fully expanded flow,and is an optimal condition for propulsive convergent nozzles, because it maximizes themomentum thrust component and therefore the net thrust.
This trend continues until point C is reached, where the fluid exiting the nozzle isequal to the velocity of sound (Me = 1.0). As illustrated in Figure 1.17, flow through aconvergent nozzle cannot be accelerated from subsonic velocities to supersonic veloci-ties, therefore as the back pressure continues to decrease past point C (to point D andbeyond) no additional mass can flow through the nozzle. (For a physical description ofthis phenomenon, see [4].) This is called choked flow. For choked flow, the exit staticpressure is not equal to the back pressure (Pe �= P0). Under these conditions the sonicgases will dissipate through a shock system. If the pressure differences are large enough,these shocks will form outside the nozzle.
36 Aerospace Propulsion Systems
Decreasing
Chokedflow
A1.0
Pc P0
P0
Pc
P0
B
e
C
D
m•
m•
Figure 1.19 Choked flow in a convergent nozzle
Choked (or underexpanded) flow occurs when the pressure ratio (Pte /P0 ) is greater thanor equal to a critical value (PRcrit). This places a maximum limit on the gas mass flowpassing through the nozzle. As previously stated, optimum (or maximum) thrust occurswhen the exhaust gases fully expand to the ambient pressure (Pe = P0). Full expansionmaximizes the momentum thrust, which maximizes the net thrust (FN ). Choked flowresults in a loss of momentum thrust, but creates a smaller pressure thrust componentsince (Pe > P0). This lost momentum thrust may only be recovered by adding a divergentsurface (e.g., a condi nozzle).
Slow down giving himmore pies Sam!
He can’t eat any faster!He is choking!
But Iwant himto win!
‘‘Choking!’’ Choked flow means that the mass flow rate has reached
a maximum value.
Fundamentals 37
For isentropic flow, Equation 1.107 can be used to derive a critical pressure ratio (PRcrit)that is necessary to just choke the nozzle, so it is the maximum pressure ratio (Pt /P ) thatcan be achieved in the nozzle. This will occur when Me = 1.0 (for a convergent nozzle)and is defined in Equation 1.145.
Critical pressure ratio:
PRcrit = Pt
P=(
1 + γ − 1
2
) γ
γ−1(1.145)
The critical pressure ratio is a function of γ only (e.g., for γ = 1.333 then PRcrit = 1.852and if γ = 1.4 then PRcrit = 1.893). Equation 1.145 provides a test to see if a convergentnozzle is choked or not. If the ideal pressure ratio (achieved by full expansion of theflow through the nozzle) exceeds the critical pressure ratio than this ideal ratio cannot beachieved because flow through the nozzle is choked.
Nozzle choke test:
Choked if:Pte
P0≥ PRcrit (1.146)
If the nozzle is choked, then the exhaust pressure ratio (Pte /Pe ) is equal to the maximumor critical pressure ratio (PRcrit). Therefore the static pressure of the exhaust gases is:
Pe = Pte
(Pe
Pte
)= Pte
PRcrit(1.147)
If the nozzle is not choked, flow is subsonic throughout the nozzle (Me < 1). Flow throughthe nozzle can adjust to changes in ambient back pressure (altitude). Ambient pressurechanges will propagate upstream from the nozzle exhaust plane at the speed of sound.So for all unchoked flows in a convergent nozzle, the exit pressure will be equal to theambient back pressure (Pe = P0). The modes of operation of a convergent nozzle aresummarized in Table 1.2.
Also from Equation 1.106, the critical temperature ratio can be found for chokedflow as:
TRcrit = γ + 1
2(1.148)
The critical temperature ratio is a function of γ only (e.g., for γ = 1.333 then TRcrit =1.167 and if γ = 1.4 then TRcrit = 1.2). The exit static temperature for a choked nozzlecan be determined from the critical temperature ratio:
Te = Tt
TRcrit(1.149)
38 Aerospace Propulsion Systems
Table 1.2 Modes of operation – convergent nozzle
Underexpanded Just choked Fully expanded(choked) (not choked)
Pte/P0 >PRcrit PRcrit <PRcrit
Pte/Pe PRcrit PRcrit <PRcrit
Me 1 1 <1Pe >P0 P0 P0
Ae(Pe − P0) >0 0 0
1.5.4.2 Convergent-Divergent (Condi) Nozzles
As was done for convergent nozzles in Figure 1.19, isentropic flow through a condi nozzlecan be understood by examining a nozzle with a constant chamber pressure (Pc) and apply-ing decreasing ambient back pressures (P0) (points A → D) on it as shown in Figure 1.20.Again Point A illustrates a limiting case where the ambient back pressure equals the cham-ber pressure (P0 = Pc), so there is no mass flow through the nozzle. Similar to Figure 1.19,as the ambient back pressure is lowered to point B and beyond (P0 < Pc), the staticpressure through the nozzle decreases and the mass flow through the nozzle increases.
In this range of ambient back pressures, the flow is fully expanded so the static pressureof the flow exiting the nozzle is equal to the ambient back pressure (Pe = P0). Flow inboth the convergent and divergent portions of the nozzle is subsonic. This trend contin-ues until the ambient back pressure at point C is reached. At this point, the fluid at thethroat flows at the velocity of sound (M∗ = 1.0). Since the flow through the convergentportion of the nozzle cannot be accelerated from subsonic velocities to supersonic veloc-ities (Figure 1.17); the condi nozzle becomes choked at all pressure ratios below point C(point D).
Decreasing
Chokedflow
eBA
1.0
C
D
Pc P0
P0
P0
Pc
m•
m•
Figure 1.20 Choked flow in a condi nozzle
Fundamentals 39
When M∗ = 1.0 at the throat, there are two possible isentropic solutions for a given arearatio (A
/A∗ ). The flow can either decelerate to a subsonic exit Mach number (Me < 1) or
accelerate to a supersonic exit Mach number (Me > 1). Point D (and lower) represents theambient back pressure condition where the flow accelerates to a supersonic Mach numberin the diverging section of the nozzle. Therefore for ambient back pressures lower thanthe point D, the pressure will decrease in both the convergent and divergent sectionsof the condi nozzle resulting in supersonic exhaust flow. This is the objective of condinozzle designs, because a supersonic exhaust gas velocity greatly increases the thrust of apropulsion system. For back pressures in between points C and D, an isentropic solutionis not possible because shock waves are formed and this is an irreversible process. In thiscase, shock equations would have to be used to determine the flow properties.
As just stated, condi nozzles are designed to be choked at the throat (M∗ = 1.0) soexhaust gases can be accelerated to a supersonic exit velocity in the diverging section.The same choke test derived for convergent nozzles (Equation 1.146) can also be appliedto the throat section of condi nozzles (assuming that the diverging section does becomea subsonic diffuser). Optimal thrust occurs when the nozzle is sized so that the exhaustgases are fully (or perfectly) expanded (Pe = P0). Imperfect nozzle expansion is causedby not having an ideal nozzle expansion ratio (ε) for a particular operating altitude.The flow is underexpanded if Pe > P0 and overexpanded if Pe < P0. Underexpansion iscaused by a less than optimal nozzle expansion ratio, resulting in a loss in momentumthrust. Overexpansion is caused by having a greater than optimal nozzle expansion ratio,which may result in flow separation, which forms shocks inside the nozzle. Nozzle per-formance losses due to overexpanded flow are generally much larger than losses due tounderexpanded flow [5].
Full expansion of an exhaust jet in a fixed-geometry condi nozzle can only be achievedwhen it is operating at its design pressure ratio. Consequently, fixed-geometry condinozzles are typically only used in missiles that spend the majority of their flight at apredictable constant supersonic cruising velocity. Most other aerospace propulsion systemsequipped with condi nozzles are designed with variable geometry (VG). This allows thearea ratio to be variably optimized over a range of flight conditions, improving the condinozzle’s effectiveness at generating thrust. A variable area nozzle is critically importantat aircraft speeds below the design speed, where severe losses can occur in the divergentsection (of the condi nozzle) due to overexpansion. Additionally most supersonic aircraftrequire an afterburner to achieve supersonic velocities. Afterburning engines are designedso that the operating conditions upstream are unchanged, which necessitates the use of avariable geometry nozzle.
Variable geometry nozzles have physical limitations that stop them from being ableto fully expand the exhaust gases at all the possible flight conditions required of anaircraft. When perfect expansion is not achieved the exhaust flow will adjust outside thenozzle by forming expansion or compression waves. Figure 1.21 shows how this occursat different conditions.
In an overexpanded condition, since the flow inside the nozzle is less than the ambientpressure (Pe< P0), oblique shocks (compression waves) form at the nozzle exit to raisethe pressure to the ambient value. The flow at the exit plane is assumed to be uniform andparallel, so by symmetry there should be no flow across the centerline. In other words,there is no velocity component normal to the centerline. The pressure of the exhaust
gases is raised to the ambient pressure as the flow goes through the first set of shocks.However, the flow in this region is turned away from the centerline. Since there canbe no normal velocity component at the centerline, the flow must turn back toward thehorizontal. Therefore the intersection of the shock waves at the centerline reflects anotherset of shock waves. As the flow passes through this second set of shock waves the pressureof the exhaust gases rises above the ambient pressure, which causes a set of expansionwaves to reflect from the ambient air. The expansion waves cause the pressure of theflow to once again be equal to the ambient pressure, but the flow is turned away from thecenterline. The intersection of the expansion waves at the centerline requires another setof expansion waves to turn the flow back towards the horizontal. These expansion wavesthen reflect from the ambient air as shock waves. This cycle continues to repeat itself untilthe exhaust gases completely mix and dissipate into the ambient air at the jet boundaries.
In an underexpanded condition the flow behaves oppositely to the overexpanded case.Since the underexpanded flow inside the nozzle was unable to decrease to the ambientpressure (Pe > P0), expansion fans form at the nozzle exit plane to reduce the pressureto the ambient value. The flow at the exit plane is assumed to be uniform and parallel,so by symmetry there should be no flow across the centerline. In other words, there is novelocity component normal to the centerline. The pressure of the exhaust gases is reducedto the ambient pressure as the flow goes through the first expansion wave. However, theflow in this region is turned away from the centerline. Since there can be no normalvelocity component at the centerline, the flow must turn back toward the horizontal there.Therefore the intersection of the expansion waves at the centerline reflects another setof expansion waves. As the flow passes through this second set of expansion waves the
Fundamentals 41
Figure 1.22 SR-71B with shock diamonds in the exhaust (Courtesy of NASA.)
pressure of the exhaust gases is reduced below the ambient pressure, which causes theexpansion waves to reflect from the ambient air as a set of oblique shocks (or compressionwaves). The shocks cause the pressure of the flow to once again be equal to the ambientpressure, but the flow is turned away from the centerline. The intersection of the shocksrequires another set of oblique shocks to turn the flow back towards the horizontal. Theseshocks then reflect from the ambient air as expansion waves. This cycle also continuesto repeat itself until the exhaust gases dissipate into the ambient air [4].
In both overexpanded and underexpanded conditions, the flow pattern behind the nozzleappears as a series of diamonds. This is often visible, as shown in Figure 1.22.
1.6 Shock Waves
A body traveling at a subsonic speed (M < 1) through a compressible fluid (such asair) creates a disturbance that is propagated throughout the fluid by a wave traveling atthe local velocity of sound (relative to the body). This creates gradual changes in thefluid properties (such as density, pressure, and temperature) along smooth, continuousstreamlines as it approaches the body. However, if the body is traveling at a supersonicspeed (M > 1) then the fluid is unable to gradually change ahead of the body. Thereforethe supersonic body induces a sudden change in fluid properties due to a shock wave.
Ahhh! I didn’t see that bump in time
I better slow down.The sign says thereis a bump ahead.
‘‘Shocking Encounter’’ The first panel shows a situation similar to super-
sonic flow over a body. Like the driver, the flow has no time to prepare for it.
The second case is like subsonic flow over a body. Like the driver, the flow
ahead of the body gradually changes to prepare for it.
42 Aerospace Propulsion Systems
Normal shock
1
P1T1V1r1
P1T1V1r1
P2T2V2r2
P2T2V2r2
1
2
2
Oblique shock
Figure 1.23 Types of shock waves
Consideration of shock waves is important in the design of intakes, nozzles, and ductsof aerospace propulsion systems capable of supersonic velocities. There are two types ofshock waves (shown in Figure 1.23). The simplest type of shock, the normal shock (orplane shock), occurs normal to the flow direction. Therefore changes to the flow propertiesacross a normal shock occur only in the flow direction. An oblique shock occurs at aninclined angle to the flow direction. The net change in fluid properties across a shock wavecan be determined by encompassing the shock within a control volume. Since there areno temperature gradients inside the control volume the shock process is adiabatic. There-fore, there is no change in the total temperature (Tt ) across the shock (Equation 1.150).However, there is a change in Mach number (velocity), static temperature (T ), total andstatic pressure (Pt and P ), and entropy (S) across the shock. So although flow through ashock wave is adiabatic, it is not isentropic because it is irreversible.
Tt1 = Tt2 ∼ Across a shock (1.150)
1.6.1 Normal Shocks
Suppose a normal shock occurs in a one-dimensional, steady flow. The shock is assumedto be thin, so that there is a negligible change in area across the shock. Applying theconservation of mass and linear momentum across the normal shock yields the followingequations, respectively:
ρ1V1 = ρ2V2 (1.151)
P1 + ρ1V21 = P2 + ρ2V
22 (1.152)
If the fluid medium is air (or some other gas that can be approximated as a perfect orideal gas), then Equations 1.13, 1.94, and 1.100 can substituted into Equations 1.151 and1.152 to transform them into the following equations:
P1
RT1M1
√γRT1 = P2
RT2M2
√γRT2 (1.153)
P1(γM2
1 + 1) = P2
(γM2
2 + 1)
(1.154)
Fundamentals 43
There is no change in total temperature across a shock (Equation 1.150). ThereforeEquation 1.106 can be substituted into Equation 1.150 to obtain:
T1
(1 + γ − 1
2
)M2
1 = T2
(1 + γ − 1
2
)M2
2 (1.155)
Rearranging Equation 1.155 gives:
P1
P2= M2
M1
√T1
T2(1.156)
Substituting Equation 1.154 and 1.155 into Equation 1.156 gives:
1 + γM22
1 + γM21
= M2
M1
√1 + γ − 1
2M2
2√1 + γ − 1
2M2
1
(1.157)
Rearranging this equation:
M1
1 + γM21
√1 + γ − 1
2M2
1 = M2
1 + γM22
√1 + γ − 1
2M2
2 (1.158)
Squaring both sides of Equation 1.158 gives:
M21
(1 + γ − 1
2M2
1
)(1 + γM2
1
)2 =M2
2
(1 + γ − 1
2M2
2
)(1 + γM2
2
)2 (1.159)
Rearranging Equation 1.159 into a quadratic equation gives:
M42
⎛⎜⎜⎝γ − 1
2−
γ 2M21
(1 + γ − 1
2M2
1
)(1 + γM2
1
)2⎞⎟⎟⎠
+M22
⎛⎜⎜⎝1 −
2γM21
(1 + γ − 1
2M2
1
)(1 + γM2
1
)2⎞⎟⎟⎠−
M21
(1 + γ − 1
2M2
1
)(1 + γM2
1
)2 = 0 (1.160)
Solving this quadratic equation for M2 gives:
M2 =
√√√√√√√M2
1 + 2
γ − 12γ
γ − 1M2
1 − 1(1.161)
44 Aerospace Propulsion Systems
Equation 1.161 shows that for supersonic flow (M1>1) then M2<1. This shows thatacross a normal shock the flow abruptly transitions from supersonic flow to subsonicflow. [Note: This equation also shows that if flow is subsonic (M1<1) and a shock occursthen M2>1. However, this second case is impossible because it violates the Second Lawof Thermodynamics by requiring an entropy decrease. Therefore for a shock to occur, theapproaching flow (M1) must be supersonic.]
From Equation 1.106, the following equation can be derived:
T2
T1=
⎛⎜⎝1 + γ − 1
2M2
1
1 + γ − 1
2M2
2
⎞⎟⎠ (1.162)
Substituting Equation 1.161 into this equation results in:
T2
T1=
(1 + γ − 1
2M2
1
)(2γ
γ − 1M2
1 − 1
)
M21
(2γ
γ − 1+ γ − 1
2
) (1.163)
Values of other property ratios across a normal shock, such as: P2 /P1 , Pt2 /Pt1 , andρ2/ρ1 shown in Equations 1.164, 1.165, and 1.166 are found in a similar manner. For
derivations of these equations, see [4].
P2
P1= 2γM2
1 − γ + 1
γ + 1(1.164)
Pt2
Pt1=
⎡⎢⎣
γ + 1
2M2
1
1 + γ − 1
2M2
1
⎤⎥⎦
γ
γ−1⎡⎢⎢⎣ 1
2γ
γ + 1M2
1 − γ − 1
γ + 1
⎤⎥⎥⎦
1γ−1
(1.165)
ρ2
ρ1= (γ + 1) M2
1
(γ − 1) M21 + 2
(1.166)
These property ratios are tabulated in Tables C.1 and C.2 (Normal Shock Tables) inAppendix C [6].
Example 1.4
A normal shock forms on the intake of an aircraft flying at Mach 1.6 at 10 km(Figure 1.24). Assume γ = 1.4. Determine the Mach number (M2), total pressure(Pt2), static pressure (P2), total temperature (Tt2), and static temperature (T2) of airafter the shock.
Fundamentals 45
Normal shock
1 2 M2
M1 = 1.6
Figure 1.24 Pitot intake
Solution
According to the Standard Atmospheric Table, Appendix A, Table A.1 for10 km altitude:
P1 = 26.5 kPa and T1 = 223.3 K
According to Table C.1, Appendix C (γ = 1.4) for M1 = 1.6:
M2 = 0.6684; Pt2
Pt1= 0.8952; P2
P1= 2.820; and
T2
T1= 1.388
Assuming isentropic flow outside the intake:
Pt1 = P1
[1 + M2
1 (γ − 1)
2
] γ
γ−1
= (26.5×103 Pa)
[1 + 1.62(1.4 − 1)
2
] 1.41.4−1
= 112.6 kPa
Tt1 = T1
[1 + M2
1 (γ − 1)
2
]
= (223.3 K)
[1 + 1.62(1.4 − 1)
2
]= 337.6 K
Therefore:
Pt2 = Pt1Pt2
Pt1= (112.6 kPa)(0.8952) = 100.8 kPa
P2 = P1P2
P1= (26.5 kPa)(2.82) = 74.7 kPa
T2 = T1T2
T1= (223.3 K)(1.388) = 309.9 K
Finally, since there is no change in total temperature across a shock:
Tt2 = Tt1 = 337.6 K
46 Aerospace Propulsion Systems
1.6.2 Oblique Shocks
The methodology of analyzing flow properties across oblique shocks is very similar tothat shown in section 1.6.1 for studying normal shocks. Even though an oblique shock isinclined at an angle to the flow direction it still creates an abrupt change in fluid propertiesand is adiabatic. Therefore like a normal shock, there is no change in the total temperature(Tt ) across an oblique shock. Also similar to a normal shock, the equations of mass, linearmomentum, and energy can be used to derive equations relating fluid properties acrossthe shock. The difference is that an additional variable must be introduced to account forthe oblique shock’s inclination to the flow direction. The two-dimensional, steady flowcontinuity equation (Equation 1.20) across an oblique shock yields the following relation:
ρ1A1V1n = ρ2A2V2n (1.167)
In Equation 1.167, V1n and V2n are the normal velocity components as shown inFigure 1.25. Since V1n and V2n are the normal components, this is essentially the samerelation that was derived for a normal shock (Equation 1.151).
The two-dimensional, steady flow equations for linear momentum (Equation 1.23) canbe separately written in component form (since momentum is a vector). This means thatmomentum can be written for both the tangential and normal directions with respect to theshock wave. Since there is no change in pressure in the tangential direction, the tangentialmomentum equation is:
V1t (ρ1A1V1n) = V2t (ρ2A2V2n) (1.168)
Equation 1.167 shows that the mass flow in the normal direction does not change acrossthe shock, therefore applying this to Equation 1.168 allows terms to be cancelled and thefollowing relation is obtained:
V1t = V2t (1.169)
This equation shows that across an oblique shock, there is no change in the tangentialvelocity. The normal momentum equation is:
ρ1A1V21n + P1A1 = ρ2A2V
22n + P2A2 (1.170)
1P1
T1 V1t V1nV2t
V2n
V2
V1
r1
P2
T2
r2
q
d
2
Figure 1.25 Velocity components across an oblique shock
Fundamentals 47
Since the shock is very thin A1 = A2 and therefore:
P1 − P2 = ρ2V2
2n − ρ1V21n (1.171)
The two-dimensional, steady flow energy equation for adiabatic steady flow(Equation 1.28) is:(
h1 + V 21
2
)−(
h2 + V 22
2
)=(
h1 + V 21t + V 2
1n
2
)−(
h2 + V 22t + V 2
2n
2
)= 0 (1.172)
Since V1t = V2t (Equation 1.169), this simplifies to:
h1 − h2 = V 22n − V 2
1n
2(1.173)
Since Equations 1.172 and 1.173 do not contain tangential velocity components, theseequations are essentially the same as the equations derived for a normal shock. Thismeans that the components normal to an oblique shock act just like a normal shock,while the components tangential to an oblique shock do not change. Therefore, the fluidproperty ratios across an oblique shock can be determined by calculating the componentsnormal to the oblique shock and using the normal shock tables found in Tables C.1 orC.2 (Appendix C).
However, oblique shock tables (Figures C.1 and C.2, Appendix C) have also been madefor the usual case, when the wave angle (δ) is unknown [6]. In order to use these tables,it is more convenient to write the components of Mach number as:
M1n = M1 sin θ (1.174)
M1t = M1 cos θ (1.175)
M2n = M2 sin(θ − δ) (1.176)
M2t = M2 cos(θ − δ) (1.177)
Figures C.1 and C.2 (Appendix C) both show that there are two possible solutions ornone at all. The three oblique shock types associated with these conditions are known as:strong shocks, weak shocks, or detached shocks.
A strong shock has a large value of θ and a large pressure ratio across the shock. Itgenerally occurs when the downstream pressure (or back pressure) of a supersonic flowis extremely high. A strong shock can be expected to occur on the spike of supersonicinlet if no flow is allowed to pass through the inlet. (However, this system is unstableand will normally degenerate into the weaker solution.) A strong shock will always slowthe supersonic flow velocity to a subsonic speed. The limiting case of a strong shock isa normal shock.
A weak shock is one that has a relatively small value of θ , a smaller pressure ratioacross the shock, and a small back pressure. A weak solution occurs more frequently
48 Aerospace Propulsion Systems
Detached shock
M > 1
M < 1M > 1
M > 1
Figure 1.26 Detached oblique shock
on aerospace system designs than a strong shock. Normally, weak shocks will occur onwings, open inlets, and planar surfaces. A weak shock will always slow the flow velocityto a lower but still supersonic speed. The limiting case of a weak shock is isentropicflow (δ = 0).
A third possibility is that there is no solution at all. This can occur if there is a greatenough wedge angle (δ). In this case the shock detaches from the body and may occurin front of it. An example of a detached bow shock is shown in Figure 1.26.
Example 1.5
Compare the loss in total pressure ratio incurred by a two-dimensional, two-shockspike diffuser and a three-shock diffuser operating at Mach 2.0, as shown inFigure 1.27. Assume that each oblique shock turns the flow through an angle (δ)of 10◦.
Two-shock inlet
1 2 3 1 2 3 4
Three-shock inlet
M1 = 2 M1 = 2
Figure 1.27
Fundamentals 49
Solution
(a) Two-shock inlet calculations:From oblique flow charts (Figures C.1a and b, Appendix C) for M1 = 2.0,γ = 1.4, and δ = 10◦, the weak shock solution is:
θ = 39.4◦
M2 = 1.64Therefore:
M1n = M1 sin θ
= (2.0) sin(39.4◦) = 1.27
The normal shock tables (Table C.1, Appendix C) can now be used forM1n = 1.27: (
Pt2
Pt1
)= 0.9842
For the normal shock M2 = 1.64, then again from the normal shock tables:(Pt3
Pt2
)= 0.8799
So the total pressure recovery across the one-shock inlet is:(Pt3
Pt1
)2 shock
inlet
= Pt3
Pt2
Pt2
Pt1= (0.8779)(0.9842) = 0.864
(b) Three-shock inlet calculations:This is done similarly to the one-shock inlet. From the oblique shock tables(Figures C.1a and b, Appendix C) again for M1 = 2.0, γ = 1.4, and δ = 10◦:
θ = 39.4◦
M2 = 1.64Therefore, once again:
M1n = M1 sin θ = 1.27
Again using the normal shock tables for M1n = 1.27:(Pt2
Pt1
)= 0.9842
For the second oblique shock for M2 = 1.64, γ = 1.4, and δ = 10◦ (Figures C.1aand b, Appendix C), θ = 49.4◦ and M3 = 1.28.
M2n = 1.64 sin(49.5◦) = 1.25
50 Aerospace Propulsion Systems
Again using the normal shock tables for M2n = 1.25:(Pt3
Pt2
)= 0.9871
For the normal shock, using the normal shock tables for M3 = 1.28:(Pt4
Pt3
)= 0.9827
Therefore: (Pt4
Pt1
)3 shock
inlet
= Pt4
Pt3
Pt3
Pt2
Pt2
Pt1
= (0.9827)(0.9871)(0.9842) = 0.9547
Thus there is about a 10 % improvement in total pressure ratio gained by usingthe three-shock inlet over a two-shock inlet at M1 = 2.0. If we were to repeat thiscalculation for M1 = 4.0, there would be a 62 % improvement. Thus the improvementincreases with higher speeds.
Example 1.5 illustrates a supersonic intake design feature for gas turbine engines andramjets. At supersonic speeds, a shock wave or series of shock waves will form at theintake slowing the airflow to a subsonic speed inside the engine. As already shown,normal shocks decrease the airflow from supersonic to subsonic speeds, while weakoblique shocks only decrease the velocity to lower supersonic speeds. Therefore for theseinlet designs, a normal shock will generally occur after an oblique shock (or a series ofoblique shocks), so that flow entering the engine will be subsonic. However, there is aloss of pressure across each shock. To increase the performance of the engine this pressureloss must be minimized. A much larger pressure loss occurs across normal shocks thanacross oblique shocks. As the supersonic speed of the flow increases, the pressure lossacross a normal shock increases exponentially, creating a stronger normal shock. Thispressure loss can be reduced by slowing the flow velocity ahead of the normal shock witha weak oblique shock (or a series of weak oblique shocks). Weak oblique shocks can beinduced ahead of the normal shock by appropriately designing the inlet geometry.
1.6.3 Conical Shocks
Supersonic flow about a three-dimensional circular cone is more complex than a simpletwo-dimensional wedge, because after a conical shock the streamlines curve to satisfy theconservation of mass. Therefore a conical shock will be inclined at a lesser angle to the flowdirection than a simple two-dimensional oblique shock. This means that a two-dimensionalwedge will create a greater flow disturbance than a three-dimensional cone. This is becauseflow cannot pass around the side of a two-dimensional wedge, since it extends to infinityin the third dimension. Therefore separate flow relations are necessary to analyze a conicalshock (see Example 5.1). These relations are illustrated in Figures C.3, C.4, and C.5 inAppendix C.
Fundamentals 51
1.7 Summary
This chapter introduced some fundamental definitions and terminology that are used in theanalysis of aerospace propulsion systems. The equations of mass, linear momentum, andenergy were presented and applied to basic engine components. Equations for isentropicflow were derived and applied to idealized engine model components. The effect ofMach number on isentropic flow was also derived. Two different types of nozzles wereintroduced: convergent and convergent–divergent (condi) nozzles. A limiting factor onthe utility of nozzles is choked flow. Choked flow means that no additional mass canflow through the nozzle (maximum mass flow). If a convergent nozzle is choked, the exitMach number of the exhaust gases is equal to 1.0 (sonic). If a condi nozzle is choked, theMach number of gases at the throat is equal to 1.0. Lastly the formation of shock wavesin compressible, supersonic flow was introduced. A normal shock represents an abruptchange in fluid properties in the direction of the flow. Although the shock is adiabatic,internal viscous dissipation and heat transfer effects make this an irreversible process.Therefore according to the Second Law of Thermodynamics, entropy will rise across ashock. This means that the flow ahead of a shock must be supersonic. Solutions to flowproperties across a normal shock were derived. Shocks that are inclined at an angle tothe flow direction are called oblique shocks. It was demonstrated that an oblique shockcan be treated as a normal shock in respect to the velocity component perpendicular tothe shock wave. Using this approach, the properties of an oblique shock can be analyzedusing the equations derived for a normal shock.
References[1] US Standard Atmosphere 1976; NOAA, NASA, USAF, Washington DC, October 1976.[2] Cengel, Y. A. and Boles, M. A., Thermodynamics, An Engineering Approach , 5th edition, McGraw-Hill,
New York, NY, USA, 2006.[3] Farokhi, S., Aircraft Propulsion , John Wiley and Sons Inc., Hoboken, NJ, USA, 2008.[4] John, J. E. A., Gas Dynamics , 2nd edition, Allyn and Bacon Inc., Newton, MA, USA, 1984.[5] Gamble, E., Terrell, D., and DeFrancesco, R., Nozzle Selection and Design Criteria, 40th
AIAA/ASME/SAE/ASEE Joint Propulsion Conference, AIAA-2004–3923, Fort Lauderdale, FLA, USA,2004.
[6] NASA Ames Research Staff, Equation, Tables and Charts for Compressible Flow , NACA Report 1135,1953.
Problems
1.1 Define the term propulsion and relate it to Newton’s Third Law of Motion.
1.2 Give a brief definition of a system. Explain the differences between open, closed,and isolated systems.
1.3 Briefly describe a working fluid. List some properties used to describe a workingfluid.
1.4 Describe how the pressure, temperature, density, and speed of sound of atmosphericair vary with altitude.
1.5 Describe the difference between a cyclic process and a non-cyclic process.
52 Aerospace Propulsion Systems
1.6 Explain the difference between a reversible process and an irreversible process.
1.7 Briefly describe what characterizes an isentropic process.
1.8 Define work. Explain how it is different from power.
1.9 Define internal energy. Describe sensible, latent, and thermal energy.
1.10 Briefly describe what is meant by the term heat. Describe the three mechanismsby which heat can be transferred.
1.11 Define the thermal efficiency of a heat engine.
1.12 Define a perfect gas. Describe how different properties of a perfect gas, such asdensity, volume, mass, temperature, and pressure, are related to one another.
1.13 Describe the continuity equation. In steady flow systems, describe what the conti-nuity equation shows about the variation of mass flow through the system.
1.14 Describe the conservation of linear momentum. If steady flow conditions exist,describe how the momentum changes with time in a constant control volume.
1.15 Describe the process of a convergent and condi nozzle becoming choked.
1.16 Explain why shock waves form when a supersonic, compressible fluid flows overa body such as a wing.
1.17 Describe the difference between normal shocks and oblique shocks. State whichshock induces the greater loss of pressure.
1.18 Briefly describe what happens to the total temperature of a fluid across a shock.
1.19 A cylinder contains 1 kg of fluid at a pressure of 100 kPa (point 1 on Figure P1.19).The fluid drives a piston by undergoing a reversible expansion defined by theequation P –V 2 = 4, until its initial volume (–V1) of 0.2 m3 doubles to a volume of(–V2) 0.4 m3. The fluid is then cooled reversibly at a constant pressure of 25 kPa untilthe piston reaches its initial position (point 3). Heat is then added reversibly withthe piston fixed until the pressure once again reaches 100 kPa (point 1). Calculatethe net work done by the fluid.
0.2
253
2
1100
P (kPa)
0.4
P 2 = 4
(m3)
Figure P1.19
1.20 A clown uses a 0.25 m3 tank containing helium (He) to fill balloons. She normallyfills the tank so that the pressure of the helium is 200 kPa at 25◦C. However, in
Fundamentals 53
order to meet the expected demands of excited children at a large party who allwant balloons, she needs to pump an additional 1 kg of helium into the tank. Sheis a little worried because she knows that the tank will rupture at a pressure of500 kPa. Helium can be assumed to be a perfect gas and has a molecular weight(Mw) of 4.003 kg/kmole. Calculate the new pressure after allowing the gas to returnto an ambient temperature of 25◦C. Will the tank rupture and ruin the party?
1.21 Oxygen (O2) at 10 MPa is stored in a pressurized, spherical tank at a temperatureof 25◦C. The tank has a radius of 25 cm. The maximum allowable pressure of thetank is rated at being 12 MPa. Assuming that oxygen is a perfect gas, calculate themass stored in the tank. To what temperature can the oxygen be allowed to risebefore the pressure limit on the tank is reached?
1.22 Calculate the specific internal energy (u) of 1 kg of air that occupies a volumeof 1 m3 at 87 kPa. If the temperature is then increased to 500 K as the air iscompressed to 200 kPa, calculate the change in internal energy and the new volumeoccupied by the air. Assume Cv = 0.718 kJ/(kg·K), Cp = 1.008 kJ/(kg·K), and R =287 J/(kg·K).
1.23 Jet fuel steadily flows through a 2 cm diameter pipe at 2 m/s. Assuming that thejet fuel is incompressible, determine the velocity (V2) of the fuel after the pipeenlarges to a 3 cm diameter (as shown in Figure P1.23).
3 cm V22 cm2 m/s
Figure P1.23
1.24 A rocket is fired on a static test stand which holds it in place. Exhaust gases areexpelled out of the rocket’s nozzle at a velocity (Vjet ) of 1500 m/s (as shown inFigure P1.24). The exhaust nozzle has a 50 cm diameter circular cross-section. Theexhaust gas has a density of 0.5 kg/m3. Determine the approximate force (or thrust)generated by the rocket, assuming the exhaust pressure equals the ambient pressure(full expansion condition).
50 cm
Vjet = 1500 m/sFN
Figure P1.24
1.25 Air flows through a compressor at a steady mass flow rate of 0.5 kg/s. The airenters the compressor at a velocity of 15 m/s, pressure of 101.3 kPa, and specificvolume of 0.8 m3/kg. The air exits at a velocity of 10 m/s, pressure of 800 kPa,and specific volume of 0.15 m3/kg. The internal energy of the air is increased by
54 Aerospace Propulsion Systems
80 kJ/kg. Calculate the power required to drive the compressor, assuming that thereis no heat loss due to a cooling system.
1.26 Air flows through a turbine at a steady mass flow rate of 15 kg/s. The air entersthe turbine at a velocity of 50 m/s and specific enthalpy (h) of 1200 kJ/kg and exitsthe turbine at a velocity of 140 m/s and specific enthalpy of 300 kJ/kg. Assumingthat no heat is rejected from the turbine, calculate the total power generated by theturbine.
1.27 An airplane flies at a velocity of 300 m/s at an altitude of 5 km. Assuming Cp =1.007 kJ/(kg·K) for air, determine the total temperature of the air relative to theairplane.
1.28 A large commercial jet aircraft is flying at Mach 0.8 at an altitude of 10 km. Airentering the intake of the aircraft’s engine is slowed down to a velocity of 100 m/s.Assume that γ = 1.4 and Cp = 1.007 kJ/(kg·K) for air. If this is an isentropicprocess, determine the static temperature and pressure of the air in the intake.
1.29 Exhaust gases exiting the nozzle of a turbojet engine (shown in Figure P1.29) have:a total temperature of 1000 K, total pressure of 350 kPa, and Mach number of 1.0.Assume isentropic flow and γ = 1.33 and R = 287 J(kg·K). Calculate the staticpressure, static temperature, and exhaust jet velocity of the gases.
Tt = 1000 K Pt = 350 kPa M = 1.0
Figure P1.29
1.30 A compressor has a total pressure ratio of 12:1. Air with a steady mass flow rate of50 kg/s enters the compressor at 500 K. If γ = 1.4, Cp = 1.007 kJ/(kg·K), and theair flows isentropically, calculate the total power required to drive this compressor.
1.31 Air with a static temperature of 223 K entering a gas turbine engine intake atV1 = 300 m/s accelerates to a new velocity (V2) and decreases in pressure (P2) atthe exit plane of the intake. The pressure recovery (P2 /P1 ) through the intake is0.833. Assume that the flow through the intake is isentropic.
(a) Calculate the static temperature (T2) of air exiting the intake.(b) Find the difference in total temperature (�Tt ) across the intake.(c) Determine the velocity (V2) of the air exiting the intake.
1.32 A turbojet engine operates at the conditions shown in the T–S diagram (Figure P1.32).Air entering the compressor is at an ambient static temperature (T1) of 200 K. Assumeisentropic diffusion through the inlet and that γ = 1.4 for the air before it enters thecombustion chamber.
(a) Determine the Mach number (M1) of the air in the inlet.(b) Determine the total pressure ratio (Pt2 /Pt1 ) across the compressor.
Fundamentals 55
0
2501
2
4
S
3
500530
1200
Tt(K)
Inta
ke
Com
pres
sor
Com
bust
ion
cham
ber
Tur
bine
Jet p
ipe
Noz
zle
1 2 3 4 65
Figure P1.32
1.33 Exhaust gases entering a convergent nozzle have a total pressure (Pt ) of 200 kPaand total temperature (Tt ) of 800 K. The gases exit the nozzle into ambient air ata static pressure (P0) of 101.3 kPa.
(a) Assuming that γ = 1.33 and R = 287 J/(kg·K), determine the critical pressureratio (PRcrit) and evaluate whether the nozzle is choked or not.
(b) Calculate the exit static pressure (Pe) and exit velocity (Ve).
1.34 A supersonic aircraft flying in air (γ = 1.4) at Mach 1.8 has an intake type whichinduces a single normal shock. Calculate the percentage pressure loss and Machnumber of the flow entering the intake diffuser after the shock.
1.35 Supersonic air (γ = 1.4) at Mach 2.8 flows over a wedge that is inclined at an angleof 30◦. If the ambient pressure is 101.3 kPa and temperature is 25◦C. Calculate theMach number, static pressure, and static temperature after the oblique shock.
1.36 A ramjet intake is designed with two ramps (Figure P1.36) so that two obliqueshocks and one normal shock occur when it travels at its cruising velocity of Mach2.8. Calculate the total pressure recovery (Pt4/Pt1) and the Mach number (M4)after this shock system. [Assume γ = 1.4 for air.]
56 Aerospace Propulsion Systems
10°
20°
M = 2.8
1
2
3
4
Figure P1.36
1.37 An attached conical shock wave forms on the nose cone of a rocket traveling atMach 8.0 at an altitude of 30 km. The cone’s semi-vertex angle (θ ) is 15◦. Calculatethe Mach number (Mc) and static pressure (Pc) of the airflow on the surface of thecone after the conical shock.
